Clarification on finite multiply transitive permutation groups and bounds on trace of associated permutation matrices Given a group $G$ such that $G < S_n$, $G \neq A_n$, and $G$ acts faithfully and transitively on the set $\Omega_n$ where $n \geq 6$, is the following statement true?

Given any non-identity element $g \in G$ with associated permutation matrix $P_g$, is its trace bounded by $0 \leq \operatorname{Tr}(P_g) \leq 5$?

My reasoning is that any finite $k$-transitive group where $k \geq 6$ must either be $S_n$ or $A_n$. Since $G$ is neither by definition, it cannot have any non-identity group element which fixes $6$ or more disjoint points and the intersection of the maximal stabilizer subgroups of any set of $6$ or more disjoint points in $\Omega_n$ must be trivial. Therefore, in the language of permutation matrices, the trace (the count of fixed points) must be less than or equal to $5$ for each non-identity group element in $G$.
This seems like a very strong restriction on permutation matrix groups, so I’m looking to see if my logic actually works or if I’ve missed extra assumptions here.
 A: No, there can be non-identity elements fixing arbitrarily many points.  For example, for integers $m, n > 1$, consider the wreath product $S_m \wr S_n := (S_m)^n \rtimes S_n$, where $S_n$ acts by permutation of the $n$ copies of $S_m$.  This acts faithfully and transitively on the set $$X = \{(i, j):  1 \leq i \leq m, 1 \leq j \leq n\}$$ of $mn$ points; it is the subgroup of $S_{mn}$ that preserves the partition of $X$ into $n$ blocks (or rows) of $m$ points of the form $(-, j)$.  This has a non-identity element that fixes $mn-2$ points, namely the element that switches $(1,1)$ with $(2,1)$ and fixes everything else.
The error in your reasoning is that $k$-transitivity is about more than just $k$-point stabilizers; it requires group elements sending any ordered $k$-tuple of distinct elements to any other.  The group described above contains some elements that stabilize a lot of points, but it doesn't mix points well enough to be $2$-transitive (or primitive for that matter):  for example, there is no group element sending $(1,1)$ to itself and $(2,1)$ to $(1,2)$.
It's also worth noting that $k$-transitivity doesn't actually imply there's a non-identity element fixing any $k$ points; in fact, this is false for groups that are sharply $k$-transitive.  (In a sharply $k$-transitive permutation group, there is a unique element sending any ordered $k$-tuple to any other, and this element will sometimes necessarily be the identity.)  What it does tell you (if $k < n$) is that there exist non-identity elements stabilizing $k-1$ points:  e.g. take an element sending $(1, 2, \dots, k-1, k)$ to $(1, 2, \dots, k-1, k+1)$, which must in particular not be the identity.
